Pre-Algebra Examples

$x (x - 2) \leq x (2 x + 6)$

Step 1

Simplify $x (x - 2)$ .

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Step 1.1

Rewrite.

$0 + 0 + x (x - 2) \leq x (2 x + 6)$

Step 1.2

Simplify by adding zeros.

$x (x - 2) \leq x (2 x + 6)$

Step 1.3

Apply the distributive property.

$x \cdot x + x \cdot - 2 \leq x (2 x + 6)$

Step 1.4

Simplify the expression.

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Step 1.4.1

Multiply $x$ by $x$ .

$x^{2} + x \cdot - 2 \leq x (2 x + 6)$

Step 1.4.2

Move $- 2$ to the left of $x$ .

$x^{2} - 2 x \leq x (2 x + 6)$

Step 2

Simplify $x (2 x + 6)$ .

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Step 2.1

Simplify by multiplying through.

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Step 2.1.1

Apply the distributive property.

$x^{2} - 2 x \leq x (2 x) + x \cdot 6$

Step 2.1.2

Reorder.

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Step 2.1.2.1

Rewrite using the commutative property of multiplication.

$x^{2} - 2 x \leq 2 x \cdot x + x \cdot 6$

Step 2.1.2.2

Move $6$ to the left of $x$ .

$x^{2} - 2 x \leq 2 x \cdot x + 6 \cdot x$

Step 2.2

Multiply $x$ by $x$ by adding the exponents.

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Step 2.2.1

Move $x$ .

$x^{2} - 2 x \leq 2 (x \cdot x) + 6 \cdot x$

Step 2.2.2

Multiply $x$ by $x$ .

$x^{2} - 2 x \leq 2 x^{2} + 6 \cdot x$

$x^{2} - 2 x \leq 2 x^{2} + 6 x$

Step 3

Move all terms containing $x$ to the left side of the inequality.

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Step 3.1

Subtract $2 x^{2}$ from both sides of the inequality.

$x^{2} - 2 x - 2 x^{2} \leq 6 x$

Step 3.2

Subtract $6 x$ from both sides of the inequality.

$x^{2} - 2 x - 2 x^{2} - 6 x \leq 0$

Step 3.3

Subtract $2 x^{2}$ from $x^{2}$ .

$- x^{2} - 2 x - 6 x \leq 0$

Step 3.4

Subtract $6 x$ from $- 2 x$ .

$- x^{2} - 8 x \leq 0$

Step 4

Convert the inequality to an equation.

$- x^{2} - 8 x = 0$

Step 5

Factor $- x$ out of $- x^{2} - 8 x$ .

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Step 5.1

Factor $- x$ out of $- x^{2}$ .

$- x \cdot x - 8 x = 0$

Step 5.2

Factor $- x$ out of $- 8 x$ .

$- x \cdot x - x \cdot 8 = 0$

Step 5.3

Factor $- x$ out of $- x (x) - x (8)$ .

$- x (x + 8) = 0$

Step 6

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x = 0$

$x + 8 = 0$

Step 7

Set $x$ equal to $0$ .

$x = 0$

Step 8

Set $x + 8$ equal to $0$ and solve for $x$ .

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Step 8.1

Set $x + 8$ equal to $0$ .

$x + 8 = 0$

Step 8.2

Subtract $8$ from both sides of the equation.

$x = - 8$

Step 9

The final solution is all the values that make $- x (x + 8) = 0$ true.

$x = 0, - 8$

Step 10

Use each root to create test intervals.

$x < - 8$

$- 8 < x < 0$

$x > 0$

Step 11

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

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Step 11.1

Test a value on the interval $x < - 8$ to see if it makes the inequality true.

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Step 11.1.1

Choose a value on the interval $x < - 8$ and see if this value makes the original inequality true.

$x = - 10$

Step 11.1.2

Replace $x$ with $- 10$ in the original inequality.

$(- 10) ((- 10) - 2) \leq (- 10) (2 (- 10) + 6)$

Step 11.1.3

The left side $120$ is less than the right side $140$ , which means that the given statement is always true.

True

Step 11.2

Test a value on the interval $- 8 < x < 0$ to see if it makes the inequality true.

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Step 11.2.1

Choose a value on the interval $- 8 < x < 0$ and see if this value makes the original inequality true.

$x = - 4$

Step 11.2.2

Replace $x$ with $- 4$ in the original inequality.

$(- 4) ((- 4) - 2) \leq (- 4) (2 (- 4) + 6)$

Step 11.2.3

The left side $24$ is greater than the right side $8$ , which means that the given statement is false.

False

Step 11.3

Test a value on the interval $x > 0$ to see if it makes the inequality true.

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Step 11.3.1

Choose a value on the interval $x > 0$ and see if this value makes the original inequality true.

$x = 2$

Step 11.3.2

Replace $x$ with $2$ in the original inequality.

$(2) ((2) - 2) \leq (2) (2 (2) + 6)$

Step 11.3.3

The left side $0$ is less than the right side $20$ , which means that the given statement is always true.

True

Step 11.4

Compare the intervals to determine which ones satisfy the original inequality.

$x < - 8$ True

$- 8 < x < 0$ False

$x > 0$ True

$x < - 8$ True

$- 8 < x < 0$ False

$x > 0$ True

Step 12

The solution consists of all of the true intervals.

$x \leq - 8$ or $x \geq 0$

Step 13

The result can be shown in multiple forms.

Inequality Form:

$x \leq - 8 or x \geq 0$

Interval Notation:

$(- \infty, - 8] \cup [0, \infty)$

Step 14